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3 months ago

Electromagnetic Induction physics ncert exemplar solutions class 12th chapter two

Two coils of self inductance L₁ and L₂ are connected in series combination having mutual inductance of the coils as M. The equivalent self inductance of the combination will be:

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Vishal Baghel

Contributor-Level 10

$L_{e q} = L_{1} + L_{2} - 2 M$

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Electromagnetic Induction Physics NCERT Exemplar Solutions Class 12th Chapter Six

Given below two statements : One is labeled as Assertion (A) and other is labeled as Reason (R).

Assertion (A) : Non-polar materials do not have any permanent dipole moment.

Reason (R) : When a non-polar material is placed in an electric field, the centre of the positive charge distribution of it's individual atom or molecule coincides with the centre of the negativne charge distribution .

In the light of above statements, choose the most appropriate answer from the options given below

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Payal Gupta

Contributor-Level 10

In non-polar molecules, centre of +ve charge coincides with centre of –ve charge, Hence, net dipole moment becomes zero.

When non-polar material is placed in external field, centre of charge does not coincide, hence give non-zero moment.

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3 months ago

Electromagnetic Induction physics ncert exemplar solutions class 12th chapter two

The dimension of mutual inductance is:

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Vishal Baghel

Contributor-Level 10

$e = - M \frac{d i}{d t}$

$M = \frac{e}{d i / d t} = M L^{2} T^{- 2} A^{- 2}$

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3 months ago

Electromagnetic Induction Physics NCERT Exemplar Solutions Class 12th Chapter Six

In a coil of resistance $8 Ω,$ the magnetic flux due to an external magnetic field varies with time as $ϕ = \frac{2}{3} (9 - t^{2}) .$ The value of total heat produced in the coil, till the flux becomes zero, will be_______J.

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alok kumar singh

Contributor-Level 10

$? = \frac{2}{3} (9 - t^{2}) = 0$

t = 3 sec

$E m f = \frac{- d ?}{d t} = \frac{4 t}{3}$

$l = \frac{E m f}{R} = \frac{\frac{4}{3} t}{8} = \frac{t}{6}$

Heat = $\int l^{2} R d t = \int_{0}^{3} \frac{t^{2}}{36} * 8 d t = 2 J$

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3 months ago

Electromagnetic Induction Physics NCERT Exemplar Solutions Class 12th Chapter Six

An aeroplane, with its wings spread 10m, is flying at a speed of 180 km/h in a horizontal direction. The total intensity of earth's field at that part is 2.5 * 10^-4Wb/m² and the angle of dip is 60°. The emf induced between the tips of the plane wings will be

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Payal Gupta

Contributor-Level 10

B_v = B sin 60°

$B_{v} = 2.5 * 1 0^{- 4} * \frac{\sqrt[]{3}}{2}$

$E m f = B_{v} * v * l = 2.5 * 1 0^{- 4} * \frac{\sqrt[]{3}}{2} * 180 * \frac{5}{18} * 1 = 108.25 * 1 0^{- 3} v o l t s$

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3 months ago

Electromagnetic Induction Physics NCERT Exemplar Solutions Class 12th Chapter Six

A transformer operating at primary voltage 8 kV and secondary voltage 160V serves a load of 80 k.W. Assuming the transformer to be ideal with purly resistive load and working on unity power factor, the loads in the primary and secondary circuit would be

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Vishal Baghel

Contributor-Level 10

$V_{P} = 8 k V, V_{S} = 160 v P = 80 k W$

(Purely resistive load)

$P = V I = \frac{V_{P}^{2}}{R_{P}} = \frac{V_{S}^{2}}{R_{S}}$

$R_{S} = \frac{V_{S}^{2}}{P} = \frac{160 * 160}{80 * 1 0^{3}} = 320 * 1 0^{- 3} = 0.32 Ω$

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3 months ago

Electromagnetic Induction Physics NCERT Exemplar Solutions Class 12th Chapter Six

A coil of inductance 1H and resistance 100 $Ω$ is connected to a battery of 6.V. Determine approximately:

(A)The time elapsed before the current acquires half of its steady – state value.

(B)The energy stored in the magnetic field associated with the coil at instant 15ms after the circuit is switched on. (Given In 2 = 0.693, e3/2 = 0.25)

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Vishal Baghel

Contributor-Level 10

L = 1H, R = 100 $Ω$

$A s, i = i_{0} e^{- t / τ}$

$F o r i = \frac{i_{0}}{2} \Rightarrow \frac{i_{0}}{2} = i_{0} e^{- t / τ}$

$\Rightarrow - l n 2 = - t / τ$

$i = \frac{6}{100} e^{- \frac{15}{1 / 100}} = . 06 e^{- 1500 * 1 0^{- 3}} = 0.06 e^{- 1.5} = 0.06 * 0.25 = . 0.015 - A$

So, $U = \frac{1}{2} L i^{2} = \frac{1}{2} * 1 * {(15 * 1 0^{- 3})}^{2} = \frac{1}{2} (225) * 1 0^{- 6} = 112.5 * 1 0^{- 6} J = 0.1125 m J$

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3 months ago

Electromagnetic Induction Physics NCERT Exemplar Solutions Class 12th Chapter Six

A metallic rod of length 20 cm is placed in North-South direction and is moved at a constant speed of 20 m/s towards East. The horizontal component of the Earth's magnetic field at that place is 4 * 10^-³ T and the angle of dip is 45°. The emf induced in the rod is………….mV.

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Payal Gupta

Contributor-Level 10

$t a n δ = \frac{B_{v}}{B_{h}}$

$B v = B_{h} t a n 45$

= 4 * 10^-³

emf induced in the rod is = $B_{v} v l$

$= 4 * 1 0^{- 3} * 20 * \frac{20}{100}$

$\begin{array}{l} = 16 * 1 0^{- 3} \\ = 16 m v \end{array}$

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5 months ago

Electromagnetic Induction Class 12th

7.17 Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements, L, C and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency.

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New answer posted

5 months ago

Electromagnetic Induction Class 12th

6.17 A line charge l per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. 6.22). A uniform magnetic field extends over a circular region within the rim. It is given by,

B = – B₀ k (r $\leq$ a; a < R)

= 0 (otherwise)

What is the angular velocity of the wheel after the field is suddenly switched off?

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Payal Gupta

Contributor-Level 10

6.17 Line charge per unit length = $λ$ = $\frac{T o t a l c h a r g e}{L e n g t h}$ = $\frac{Q}{2 π r}$

Where r = distance of the point within the wheel

Mass of the wheel = M

Radius of the wheel = R

Magnetic field, $\overset{?}{B}$ = $- B_{0} \overset{?}{k}$

At a distance r, the magnetic force is balanced by the centripetal force. i.e.

BQ $v$ = $\frac{M v^{2}}{r}$ , where v = linear velocity of the wheel. Then,

B $* 2 λ π r$ = $\frac{M v}{r}$

v = $\frac{2 B λ π r^{2}}{M}$

Angular velocity, $ω = \frac{v}{R}$ = $\frac{2 B λ π r^{2}}{M R}$

For r $\leq a \leq R, w e g e t$ $ω = - \frac{2 B_{0} λ π a^{2}}{M R} \overset{?}{k}$

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